# Double Integrals Over Rectangles

Recall from the Defining the Integral and Riemann Sums that if $y = f(x)$ is a continuous (over $[a, b]$) single variable real-valued function, then the definite integral from $x = a$ to $x = b$ is defined to be:

(1)

When $f(x) ≥ 0$ $\forall x \in [a, b]$, then the definite integral from $a$ to $b$ represented the area trapped between the curve $f$, the $x$-axis, and the lines $x = a$ and $x = b$.

We will now extend the concept of integration to functions of two variables. Let $z = f(x, y)$ be a two variable real-valued function that generates a surface, and that is defined on a rectangular subset of the domain of $f$. We will call this subset of the domain $R$, and we define it to be:

(2)

Suppose that $f(x,y) ≥ 0$ for all $(x, y) \in R$, and define the solid $S_R = \{ (x, y, z) \in \mathbb{R}^3 : (x, y) \in R \}$, that is $S_R$ is the solid generated by the function $z = f(x, y)$ for $(x, y) \in R$, the $xy$-plane, and the vertical lines that pass through the points $(a, c)$, $(a, d)$, $(b, c)$, and $(b, d)$.

We will approximate this volume of $S_R$ by taking the domain and dividing it into subrectangles by dividing the interval $[a, b]$ into $m$ equal width $\Delta x = \frac{b – a}{m}$ subintervals $[x_{i-1}, x_i]$, and by diving the interval $[c, d]$ into $n$ equal width $\Delta y = \frac{d – c}{n}$ subintervals $[y_{j-1}, y_j]$. We will define the $ij^{\mathrm{th}}$ subrectangle as:

(3)

The area of each of these subrectangles is $\Delta A = \Delta x \Delta y$. Now we will take any point $(x,y) \in R_{ij}$. The set of points in $R_{ij}$ are called sample points, and the one point that we choose in specific for each subrectangle we will denote by $(x_{ij}^*, y_{ij}^*)$.

The height of each rectangular column is $f(x_{ij}^*, y_{ij}^*)$, and so the volume $V_{ij}$ of any rectangular column is given by the formula $V_{ij} = f(x_{ij}^*, y_{ij}^*) \Delta A$.

The volume of $S_R$ can thus be approximated as the sum of the volumes of all of these rectangular columns:

(4)

As the number of subrectangles increases in our approximation of the volume – we can see that the formula above approximates the actual volume more accurately, that is as $m \to \infty$, and $n \to \infty$, the approximation of the volume $V$ of $S_R$ gets better and better, so:

(5)

We will now define the double integral as follows.

Definition: Let $z = f(x, y)$ be a two variable real-valued function, and let the rectangle $R \subseteq D(f)$. Then the Double Integral of $f$ over $R$ denoted $\iint_{R} f(x, y) \: dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A$ provided that this limit exists. The function $z = f(x, y)$ is said to be Integrable over $R$ if this limit exists. |

*Note that the double integral of $f$ over $R$ does not restrict $f$ to be solely above the $xy$-plane. If $f$ is above the $xy$-plane, then the double integral of $f$ over $R$ represents the volume of the sold. If not, then the limit will not represent the volume of the solid.*

The following is a more precise definition of the double integral.

Definition (Formal): Let $z = f(x, y)$ be a two variable real-valued function, and let the rectangle $R \subseteq D(f)$. Then the Double Integral of $f$ over $R$ is defined as $\iint_{R} f(x, y) \: dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A$ if $\forall \epsilon > 0$ $\exists N \in \mathbb{Z}$ such that for all $m, n \in \mathbb{Z}$, $m, n ≥ N$ and for any choice of sample points $(x_{ij}^*, y_{ij}^*)$ we have that $\biggr \rvert \iint_{R} f(x,y) \: d A – \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A \biggr \rvert . |

*If we choose all of the sample points to be the upper right corners for each $R_{ij}$, that is, choosing the sample points to be $(x_i, y_j)$, then $\iint_{R} f(x, y) \: dA = \lim_{m,n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i, y_j) \Delta A$.*

We will now look at an example of finding the volume under a curve.

## Example 1

**Let $f(x, y) = 3x + 2y + 1$ and let $R = [0,1] \times [0, 1]$. Find the volume trapped between $R$, the surface generated by $f$ over $R$, and the vertical lines that pass through $(0,0)$, $(1,0)$, $(1,1)$ and $(0, 1)$.**

First notice that $f$ lies above the $xy$ plane on $R = [0,1] \times [0, 1]$.

We note that $\Delta x = \frac{b – a}{m} = \frac{1 – 0}{m} = \frac{1}{m}$ and $\Delta y = \frac{d – c}{n} = \frac{1-0}{n} = \frac{1}{n}$, and so $\Delta A = \Delta x \Delta y = \frac{1}{m} \frac{1}{n} = \frac{1}{mn}$.

Now we note that $x_i = i\Delta x = \frac{i}{m}$ and $y_j = j \Delta y = \frac{j}{n}$. Therefore $f(x_i, y_j) = \frac{3i}{m} + \frac{2j}{n} + 1$.

Before we compute the limit, recall that $\sum_{a=1}^{b} a = \frac{b(b+1)}{2}$, $\sum_{a=1}^{b} C = bC$ where $C$ is any term that contains no $a$‘s. We use these identities a few times in simplifying the following limit:

(6)

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